Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total…
Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is
$2^{\frac{2}{3}}:1$
$\sqrt{2}: 1$
$2^{\frac{1}{3}}: 1$
$2: 1$
Solution
The volume of mercury remains same.
$\begin{array}{l}
\therefore \frac{4 \pi}{3} R^{3}=2\left(\frac{4 \pi}{3} r^{3}\right) \\
\therefore R^{3}=2 r^{3} \quad \text { or } \quad r^{3}=\frac{R^{3}}{2} \\
\therefore r=\frac{R}{2^{\frac{1}{3}}}
\end{array}$
Initial surface energy $\mathrm{E}_{1}=4 \pi \mathrm{R}^{2} \cdot \mathrm{T}$
Final surface energy
$\mathrm{E}_{2}=4 \pi \mathrm{r}^{2} \mathrm{~T}=4 \pi\left(\frac{\mathrm{R}}{2^{\frac{1}{3}}}\right)^{2} \mathrm{~T}=4 \pi \frac{\mathrm{R}^{2}}{2^{\frac{2}{3}} \mathrm{~T}}$
$\therefore \frac{E_{1}}{E_{2}}=2^{\frac{2}{3}}$